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प्रश्न
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
विकल्प
`1/sqrt2`
`1/sqrt3`
`1/(2sqrt2)`
`1/(3sqrt3)`
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उत्तर
(c) `1/(2sqrt2)`
Let \[\sin^{- 1} \frac{\sqrt{63}}{8} = y\]
Then,
\[\sin{y} = \frac{\sqrt{63}}{8}\]
\[\cos{y} = \sqrt{1 - \sin^2 y} = \sqrt{1 - \frac{63}{64}} = \frac{1}{8}\]
Now, we have
\[\sin\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right) = \sin\left( \frac{1}{4}y \right)\]
\[ = \sqrt{\frac{1 - \cos\frac{y}{2}}{2}} \left[ \because \cos2x = 1 - 2 \sin^2 x \right]\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{1 + \cos{y}}{2}}}{2}} \left[ \because \cos2x = 2 \cos^2 x - 1 \right]\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{1 + \frac{1}{8}}{2}}}{2}}\]
\[ = \sqrt{\frac{1 - \sqrt{\frac{9}{16}}}{2}}\]
\[ = \sqrt{\frac{1 - \frac{3}{4}}{2}}\]
\[ = \sqrt{\frac{1}{8}}\]
\[ = \frac{1}{2\sqrt{2}}\]
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